【Read between lines】Hal Tasaki "Physics and mathematics of quantum many body systems" p. 44 (proof of Marshal-Lieb-Mattis theorem generalized to antiferromagnetic Heisenberg model with anisotropy)

Keywords

• Marshal-Lieb-Mattis theorem

• Antiferromagnetic Heisenberg model with anisotropy

• Perron-Frobenius theorem to a real-symmetric matrix

Solutions

Overview

Since Hamiltonian commutes with $${\hat{S}_\mathrm{tot}^{(3)}}$$, we can take simultaneous eigenstates. We use only simultaneous eigenstates, and label eigenvalues of $${\hat{S}_\mathrm{tot}^{(3)}}$$ by $${M}$$.

We should prove

1. The ground state is not degenerate.

2. The ground state is in $${M=0}$$ sector.

The proof that the ground state is not degenerate is done by contradiction. 

Roughly, we assume that the ground state is degenerate. By continuous change of parameters from the isotropic model of $${(\Lambda, D)=(1,0)}$$, the Perron-Frobenius theorem in the $${M=0}$$ sector, and $${π}$$ rotation symmetry around the $${y}$$ axis, we can show that the assumed ground states are at least threefold degenerate. This is the first paragraph.

On the other hand, for each of the cases (i) and (ii), we sandwich $${H'}$$, whose energy spectrum is identical to the original $${H}$$ but whose rotation axis of U(1) is replaced by $${z\to y}$$, with the appropriate basis. Using the Perron-Frobenius theorem by grouping $${M}$$ together in even sectors, we can show that the lowest eigenvalue state is not degenerate in the Hilbert space of $${M=0, \pm2, \pm4, ...}$$ combined. Similarly, it can be shown that the eigenstates of the lowest eigenvalue are not degenerate in the Hilbert space of $${M=0, \pm1, \pm3, ...}$$ together. These facts contradict the assumption, because the degeneracy is at most double in the whole Hilbert space, while we derived threefold degeneracy in the first paragraph. Case (i) is shown in the second paragraph. Case (ii) is the third paragraph of the proof.

$${π}$$ rotation around $${y}$$ axis (i.e. $${\exp(-iπ\hat{S}_\mathrm{tot}^{(2)})}$$) and Hamiltonian commute. This means if the state $${\ket{\Phi_M}}$$ with eigenvalue $${M}$$ is an energy eigenstate, then $${\ket{\Phi_{-M}}}$$ is also an eigenstate with the same energy eigenvalue. Only $${M=0}$$ is not degenerate.

Here we will see only the proof of no degeneracy of the ground state. We rearrange the flow of the proof:

1. Prove the minimum eigenvalue state in M≠0 sector has a degenerated state in -M sector

2. Prove that M=0 sector has a unique minimum eigenvalue state

3. Prove that, if the ground state shifts from M=0 sector to another, there is a point in the parameter space where the ground state degenerates threefold.

4. Prove that case (i) has a unique minimum eigenvalue state in union of even M's, and one in union of odd M's

5. Prove that case (ii) also has a unique minimum eigenvalue state in union of even M's, and one in union of odd M's

Step 1: The minimum eigenvalue state in M≠0 sector has a degenerated state in -M sector

First, we show that the ground state is at least doubly degenerated, if one ground state is in $${\mathscr{H}_M}$$ with $${M\neq0}$$. Suppose that $${\ket{\Phi_M}\in\mathscr{H}_M}$$  is a minimum eigenvalue state. Since $${\exp(-iπ\hat{S}_\mathrm{tot}^{(2)})}$$ commutes with Hamiltonian, the state $${\exp(-iπ\hat{S}_\mathrm{tot}^{(2)})\ket{\Phi_M}}$$ has the same energy. This state is also an eigenstate of $${\hat{S}_\mathrm{tot}^{(3)}}$$. The eigenvalue is $${-M}$$:

$$
\hat{S}_\mathrm{tot}^{(3)}\exp(-iπ\hat{S}_\mathrm{tot}^{(2)})\ket{\Phi_M}=-\exp(-iπ\hat{S}_\mathrm{tot}^{(2)})\hat{S}_\mathrm{tot}^{(3)}\ket{\Phi_M}=-M\exp(-iπ\hat{S}_\mathrm{tot}^{(2)})\ket{\Phi_M}.
$$

Therefore if the ground state is in $${M\neq0}$$ sector, there is a degenerated ground state in $${\mathscr{H}_{-M}}$$, but there may be no degeneracy if the ground state is in $${\mathscr{H}_0}$$.

Step 2: M=0 sector has an unique minimum eigenvalue

Next, we will see that the minimum eigenvalue in $${\mathscr{H}_0}$$ sector is not degenerate by Perron-Frobenius theorem. Rewrite Hamiltonian:

$$
\hat{H}=\sum_{\{x,y\}\in\mathscr{B}}\left(\frac{\hat{S}_x^+\hat{S}_y^-+\hat{S}_x^-\hat{S}_y^+}{2}+\lambda\hat{S}_x^{(3)}\hat{S}_y^{(3)}\right)+D\sum_{x\in\Lambda}(\hat{S}_x^{(3)})^2.
$$

The matrix component $${\bra{\tilde\Psi^\sigma}\hat{H}\ket{\tilde\Psi^{\sigma'}}}$$ with $${\bar{\sigma}=\bar{\sigma'}=0}$$ is real symmetric.
One of $${\{x, y\}\in\mathscr{B}}$$ is in A sub lattice, the other is in B; either changes the sign of $${\ket{\tilde\Psi^\sigma}}$$. Thus $${(\hat{S}_x^+\hat{S}_y^-+\hat{S}_x^-\hat{S}_y^+)/2}$$ contributes negative sign. When $${\sigma\neq\sigma'}$$, the remaining terms vanish, because $${\hat{S}_x^{(3)}}$$ does not change the basis $${\ket{\tilde\Psi^\sigma}}$$. Therefore $${\bra{\tilde\Psi^\sigma}\hat{H}\ket{\tilde\Psi^{\sigma'}}\leq0}$$, if off-diagonal.
The Proposition in this section ensures that the off-diagonal components are connected.
From the above, all the requirements of Perron-Frobenius theorem is satisfied in $${\mathscr{H}_0}$$ sector, thus the lowest eigenvalue is not degenerate.

From above, there can be two cases:

1. One ground state is in $${\mathscr{H}_M}$$, and another in $${\mathscr{H}_{-M}}$$, with $${M\neq0}$$. The ground states are at least doubly degenerated.

2. The unique ground state is in $${\mathscr{H}_0}$$.

Step 3: There is a threefold degenerated ground state if the ground state shifts from M=0 sector to another

Above we showed that the $${M=0}$$ sector has a unique minimum eigenvalue state, but $${M≠0}$$ does not, at any point in parameter space $${\{(\lambda, D)\in\mathbb{R}^2\}}$$. It is implicitly assumed that you can change parameters continuously in $${\mathbb{R}^2}$$.

Let $${E_0(\lambda,D), E_1(\lambda, D), \varepsilon(\lambda, D)}$$ be the minimum energy eigenvalue in $${\mathscr{H}_0}$$, and that in $${\mathscr{H}\setminus\mathscr{H}_0}$$, respectively. Define $${\varepsilon(\lambda, D):=E_1(\lambda, D)-E_0(\lambda, D)}$$. Marshal-Lieb-Mattis theorem says that there is a unique ground state in $${\mathscr{H}_0}$$ in case of $${(\lambda, D)=(1,0)}$$, i.e., $${\varepsilon(1,0)>0}$$. Assume that there exists $${(\lambda_\ast, D_\ast)\in\mathbb{R}^2}$$ such that $${\varepsilon(\lambda_\ast, D_\ast)<0}$$, which means the ground state is not in $${\mathscr{H}_0}$$ but in $${\mathscr{H}_{M\neq0}}$$. Then there is a point $${(\lambda_0, D_0)\in\mathbb{R}^2}$$ such that $${\varepsilon(\lambda_0, D_0)=0}$$ along any continuous path which connects $${(\lambda_\ast,D_\ast)}$$ and $${(1,0)}$$. At this point ground states are at least threefold degenerated: one in $${\mathscr{H}_0}$$, another is in $${\mathscr{H}_{M}}$$ for $${M\neq0}$$, and another (or the other) is in $${\mathscr{H}_{-M}}$$.

Step 4: Case (i) has unique minimum eigenvalue state in union of even M's, and one in union of odd M's

The statement for even M's is proven by Perron-Frobenius theorem for matrix $${\bra{\tilde\Psi^\sigma}\hat{H}'\ket{\tilde\Psi^{\sigma'}}}$$ with spin configuration $${\bar\sigma\in\{0, \pm2, \pm4, …\}}$$. This statement says there is a unique minimum eigenvalue state in $${…\cup\mathscr{H}_{-2}\cup\mathscr{H}_{0}\cup\mathscr{H}_2\cup…}$$. You can check requirements of Perron-Frobenius theorem as written in the textbook. Here, we verify the connectivity of off-diagonal components in the rest of this step. Note that $${\hat{H}’}$$ has the same energy spectrum with $${\hat{H}}$$, but $${\hat{H}'}$$ doesn’t commute with $${\hat{S}_\mathrm{tot}^{(3)}}$$, hence we cannot take simultaneous eigenstates.

The terms $${(1+\lambda)(\hat{S}_x^+\hat{S}_y^-+\hat{S}_x^-\hat{S}_y^+)/4}$$ of (2.5.16) has nonzero contribution when $${\sigma_x+\sigma_y=\sigma_x'+\sigma_y'}$$ is satisfied. Note that $${\hat{S}_x^{(3)}\hat{S}_y^{(3)}, -\hat{S}_x^+\hat{S}_x^-, -\hat{S}_x^-\hat{S}_x^+}$$ vanish when $${\sigma\neq\sigma'}$$. If both $${\bra{\tilde\Psi^\sigma}}$$ and $${\ket{\tilde\Psi^{\sigma'}}}$$ have a common $${\hat{S}_\mathrm{tot}^{(3)}}$$ eigenvalue, there is no nonzero term but the first. The Proposition in this section is applicable in case of $${\bar\sigma=\bar\sigma'}$$, since the situation is identical to the conditions of Marshal-Lieb-Mattis theorem. Spin configurations with $${\bar\sigma=\bar\sigma'}$$ are, therefore, connected.

We can show spin configurations with $${\bar\sigma=\bar\sigma'+2}$$ are also connected. In case $${\sigma_x=\sigma_x'+2}$$ for one site $${x\in\Lambda}$$ and $${\sigma_y=\sigma_y'}$$ for the other sites ($${y\in\Lambda\setminus\{x\}}$$), the non-vanishing term is calculated as the follows:

$$
\bra{\tilde\Psi^\sigma}\hat{H}'\ket{\tilde\Psi^{\sigma'}}=\bra{\tilde\Psi^\sigma}-\frac{D}{4}(\hat{S}_x^+)^2\ket{\tilde\Psi^{\sigma'}}.
$$

In case $${\sigma_x=\sigma_x'+1}$$ for one site $${x\in\Lambda}$$ and $${\sigma_y=\sigma_y'+1}$$ for another site $${y\in\Lambda\setminus\{x\}}$$, the non-vanishing term is,

$$
\bra{\tilde\Psi^\sigma}\hat{H}'\ket{\tilde\Psi^{\sigma'}}=\frac{1-\lambda}{4}\bra{\tilde\Psi^\sigma}\hat{S}_x^+\hat{S}_y^+\ket{\tilde\Psi^{\sigma'}}.
$$

Either of two is non-zero because $${(\lambda, D)\neq(1,0)}$$. Therefore, there is at least one connection in spin configuration with $${\bar\sigma=\bar\sigma'}$$, hence off-diagonal components in $${…, \mathscr{H}_{-2}, \mathscr{H}_{0}, \mathscr{H}_{2}, …}$$ are all connected. 

From above, all requirements of Perron-Frobenius theorem are satisfied in $${…\cup\mathscr{H}_{-2}\cup\mathscr{H}_{0}\cup\mathscr{H}_2\cup…}$$. Therefore the minimum eigenvalue state in this union of sectors is not degenerate.

For the same reason $${…\cup\mathscr{H}_{-3}\cup\mathscr{H}_{-1}\cup\mathscr{H}_1\cup\mathscr{H}_3\cup…}$$ has unique minimum eigenvalue state.

The ground state of $${\hat{H}'}$$ satisfies one of the followings:

1. Unique ground state is in $${…, \mathscr{H}_{-2}, \mathscr{H}_{0}, \mathscr{H}_{2}, …}$$.

2. Unique ground state is in $${…\cup\mathscr{H}_{-3}\cup\mathscr{H}_{-1}\cup\mathscr{H}_1\cup\mathscr{H}_3\cup…}$$.

3. Ground state is twofold: one is in $${…, \mathscr{H}_{-2}, \mathscr{H}_{0}, \mathscr{H}_{2}, …}$$; the other is in$${…\cup\mathscr{H}_{-3}\cup\mathscr{H}_{-1}\cup\mathscr{H}_1\cup\mathscr{H}_3\cup…}$$.

Therefore the ground state is at most doubly degenerated. This is satisfied also for $${\hat{H}}$$, which has identical energy spectrum.

Step 5: Case (ii) has unique minimum eigenvalue state in union of odd M, and one in union of even M

The logic is the same as step 3. We use the following state:

$$
\ket{\tilde{\tilde{\Psi}}^\sigma}=(-1)^{(\bar\sigma-|\Lambda|S)/2}\ket{\tilde\Psi^\sigma}
$$

Here we verify that off-diagonal components are non-positive. $${\hat{S}_x^\pm\hat{S}_y^\mp}$$ does not change the sign from $${\bar\sigma}$$ but one $${\sigma_x}$$ of $${\ket{\tilde\Psi}}$$. $${\hat{S}_x^\pm\hat{S}_y^\pm}$$ changes the sign from both $${\bar\sigma/2}$$ and one $${\sigma_x}$$. $${(\hat{S}_x^+)^2}$$ changes only the sign by $${\bar\sigma/2}$$. Therefore if $${\sigma\neq\sigma'}$$ and $${\{x,y\}\in\mathscr{B}}$$, 

$$
\bra{\tilde{\tilde{\Psi}}^\sigma}\hat{S}_x^\pm\hat{S}_y^\mp\ket{\tilde{\tilde\Psi}^{\sigma'}}\leq0, \quad\bra{\tilde{\tilde{\Psi}}^\sigma}\hat{S}_x^\pm\hat{S}_y^\pm\ket{\tilde{\tilde\Psi}^{\sigma'}}\geq0, \quad\bra{\tilde{\tilde{\Psi}}^\sigma}(\hat{S}_x^\pm)^2\ket{\tilde{\tilde\Psi}^{\sigma'}}\leq0.
$$

Therefore all the off-diagonal components are non-positive. The other requirements of Perron-Frobenius theorem is proven in the same way as case (i). The result is also the same: the ground state is at most doubly degenerated. This is satisfied for $${\hat{H}}$$, too.

Conclusion

Step 1 to Step 3 says there is two cases for this model:

1.  The ground state is unique and in $${\mathscr{H}_0}$$, for any $${(\lambda, D)\in((-1,1]\times[D,\infty))\cup([1,\infty)\times(-\infty,0])}$$.

2. There is a point in $${((-1,1]\times[D,\infty))\cup([1,\infty)\times(-\infty,0])}$$ where the ground state is threefold degenerated.

On the other hand, Step 4 and Step 5 says the ground state is at most doubly degenerated, hence all the statement is proven. $${\blacksquare}$$